Consider a branch of the hyperbola x² - 2y² - 2√2x - 4√2y - 6 = 0 with vertex at the point A. Let B

Q: Wait, why do we reject the negative root for the vertex? Or is it just one vertex?

Great observation! A hyperbola has two vertices. The problem asks for 'a' branch, and we are calculating the area for 'a' triangle formed by 'a' vertex and 'a' latus rectum endpoint. By symmetry, the area will be the same regardless of which branch or which end of the latus rectum you choose. We pick the positive side for simplicity, but the magnitude of the area remains invariant.

Q: Why can't we just use the coordinate geometry area formula $\frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$?

You absolutely can! It is a perfectly valid, robust method. However, in JEE Advanced, time is your most precious currency. Recognizing that $\triangle ABC$ is a right-angled triangle allows you to bypass the determinant formula and solve it in seconds. Always look for the geometric shortcut first; it is the hallmark of a master student.

Q: How do I know the latus rectum is vertical?

The orientation of the latus rectum depends on the transverse axis. Since our hyperbola is of the form $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$, the transverse axis is horizontal (parallel to the x-axis). Therefore, the latus rectum, which is perpendicular to the transverse axis, must be vertical. Always sketch the hyperbola first to confirm this!

Question

Consider a branch of the hyperbola $x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$ with vertex at the point $A$. Let $B$ be one of the end points of its latus rectum. If $C$ is the focus of the hyperbola nearest to the point $A$, then the area of the triangle $ABC$ is

Options:

(A) $1 - \sqrt{\frac{2}{3}}$

(B) $\sqrt{\frac{3}{2}} - 1$

(C) $1 + \sqrt{\frac{2}{3}}$

(D) $\sqrt{\frac{3}{2}} + 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex, second-degree equation: 
$$x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$$

It looks like a tangled mess, but in the world of coordinate geometry, this is merely a mask. Beneath this chaotic exterior lies the elegant, sweeping curve of a hyperbola. Our mission is to perform a surgical procedure on the equation to reveal its DNA.

### The Art of Completing the Square

To find the truth, we must transform this general form into the standard form:
$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$

We start by grouping the $x$ terms and the $y$ terms. For the $x$ terms, we have $x^2 - 2\sqrt{2}x$. To complete the square, we take half of the coefficient of $x$, which is $\sqrt{2}$, and square it to get $2$. By adding and subtracting $2$, we collapse this into $(x-\sqrt{2})^2 - 2$.

We apply the same rigorous logic to the $y$ terms. Factoring out the $-2$ from $-2y^2 - 4\sqrt{2}y$ gives us $-2(y^2 + 2\sqrt{2}y)$. Completing the square inside the parenthesis results in $-2(y+\sqrt{2})^2 + 4$.

When we combine these pieces with the constant $-6$, the equation simplifies to:
$$(x-\sqrt{2})^2 - 2(y+\sqrt{2})^2 = 4$$

Dividing by $4$, we arrive at our standard form:
$$\frac{(x-\sqrt{2})^2}{4} - \frac{(y+\sqrt{2})^2}{2} = 1$$

### Extracting the DNA

With the equation in standard form, the parameters $a$, $b$, and the center $(h, k)$ reveal themselves. We see $a^2 = 4$, so $a = 2$. We see $b^2 = 2$, so $b = \sqrt{2}$. The center is at $(\sqrt{2}, -\sqrt{2})$.

The eccentricity $e$ is the measure of how 'stretched' our hyperbola is. Using the formula $e = \sqrt{1 + \frac{b^2}{a^2}}$, we calculate:
$$e = \sqrt{1 + \frac{2}{4}} = \sqrt{\frac{3}{2}}$$

### Plotting the Geography

Now, let's place our points on the coordinate plane. The vertex $A$ lies on the transverse axis at $(h+a, k)$, which gives us $(\sqrt{2}+2, -\sqrt{2})$.

The focus $C$ is located at $(h+ae, k)$. Calculating $ae$, we get $2 	imes \sqrt{\frac{3}{2}} = \sqrt{6}$. Thus, $C$ is at $(\sqrt{2}+\sqrt{6}, -\sqrt{2})$.

The latus rectum passes through the focus $C$ vertically. The endpoint $B$ is at $(x_C, k + \frac{b^2}{a})$. Since $\frac{b^2}{a} = \frac{2}{2} = 1$, point $B$ is at $(\sqrt{2}+\sqrt{6}, -\sqrt{2}+1)$.

### The Geometric Revelation

Look at the triangle $ABC$. The segment $AC$ lies on the horizontal transverse axis, and the segment $BC$ is part of the vertical latus rectum. A horizontal line meeting a vertical line creates a perfect $90^\circ$ angle at $C$.

We have a right-angled triangle where the base $AC$ is the difference in $x$-coordinates:
$$AC = (\sqrt{2}+\sqrt{6}) - (\sqrt{2}+2) = \sqrt{6}-2$$

The height $BC$ is simply the semi-latus rectum length, which is $1$. The area is calculated as:
$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (\sqrt{6}-2) 	imes 1$$

Simplifying this, we get:
$$\mathbf{\frac{\sqrt{6}}{2} - 1}$$

This is the elegant truth hidden within the chaos.

Select Answer:

Visualized Solution

The Hyperbola Equation

Completing the Square for $x$

Completing the Square for $y$

Standard Form Identification

Extracting Parameters $a$ and $b$

Calculating Eccentricity $e$

Locating Vertex $A$

Locating Focus $C$

Locating Latus Rectum Endpoint $B$

Visualizing Triangle $A B C$

Calculating Base $A C$

Calculating Height $B C$

Final Area Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Art of Completing the Square

Extracting the DNA

Plotting the Geography

The Geometric Revelation

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Select Answer:

Visualized Solution

The Hyperbola Equation

Completing the Square for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Completing the Square for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Standard Form Identification

Extracting Parameters .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Calculating Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e

Locating Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Locating Focus .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Locating Latus Rectum Endpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Visualizing Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC

Calculating Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Calculating Height .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC